Homework 7 Solution (Due Nov 7th, Thursday)Problems:1) 7.4a) Using the “SS1” option in SAS “PROC REG”, we obtain the ANOVA table with the extra sumsof squares below. Note that we need to put the variables into the model in the order: y = x1x3x2.The res ult i ng values are: SSR(X1) = 136, 366, df=1; SSR(X3| X1)=2, 033, 565, df = 1; SSR(X2|X1,X3) = 6674.6, df=1.b) For this problem, we also need the SSE(X1,X2,X3) or MSE(X1,X2,X3), which can be obt ai n edfrom the ANOVA table below. SSE(X1,X2,X3) = 985, 530, and MSE(X1,X2,X3) = 985, 530/48 =20532. Here, we are testing the following: H0: β2=0vs. Ha: β2�= 0. Note that SSR(X2| X1,X3)=6674.6, SSE(X1,X2,X3) = 985, 530, F�= (6674.6/1) ÷ (985, 530/48) = 0.32491.The critical F value can be obtained from SAS using the “finv” function. We need the F value:F (0.95,df =(1, 48)) = 4. 04265. Since the F∗valu e is less than 4.04265, we do not reject the null andconclude that the slope β2= 0 . We can obtain the p-value us i ng SAS, with the command: “pf=1-probf(0.32491,1,48)”. Th e p-value is = 0.5713. Since the p-value is not less t han 0.05, we conclude thenull. Thus, X2can be dropped from the regression model given that X1and X3are retained.Analysis of VarianceSum of MeanSource DF Squares Square F Value Pr > FModel 3 2176606 725535 35.34 <.0001Error 48 985530 20532Corrected Total 51 3162136Root MSE 143.28946 R-Square 0.6883Dependent Mean 4363.03846 Adj R-Sq 0.6689Coeff Var 3.28417Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t| Type I SSIntercept 1 4149.88721 195.56541 21.22 <.0001 989877440ship 1 0.00078708 0.00036455 2.16 0.0359 136366holiday 1 623.55448 62.64095 9.95 <.0001 2033565cost 1 -13.16602 23.09173 -0.57 0.5712 6674.58809Obs fcri1 fprob11 4.04265 0.57133c) Here, we need to run twonew SAS mo d el s :y = x1 x2 / ss1;y = x2 x1 / ss1.The output is b el ow.Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t| Type I SSIntercept 1 3995.47867 337.76602 11.83 <.0001 989877440ship 1 0.00091916 0.00063120 1.46 0.1517 136366cost 1 12.12052 39.76555 0.30 0.7618 5725.92181Parameter EstimatesParameter Standard21Variable DF Estimate Error t Value Pr > |t| Type I SSIntercept 1 3995.47867 337.76602 11.83 <.0001 989877440cost 1 12.12052 39.76555 0.30 0.7618 11395ship 1 0.00091916 0.00063120 1.46 0.1517 130697Here, we see that SSR(X1)+SSR(X2| X1) = 136, 366+5726 = 142, 092. Also, SSR(X2)+SSR(X1|X2) = 11, 395 + 130, 697 = 142, 092. This is always the case, since both equal SSR(X1,X2).2) 7.6The SAS output is be l ow. Based on this, SSR(X2,X3| X1)=SSR(X2| X1)+SSR(X3| X1,X2)=480.91529 + 364.15952 = 845.07, with df=2. The SSE(X1,X2,X3) = 4248.84, with df=42, andMSE(X1,X2,X3) = 101.163.Here, we are testing H0: β2= β3=0vs. Ha: not both β2and β3= 0. SSR(X2,x3| X1) = 845.07,SSE(X1,X2,X3)=4, 248.84, F�= (845.07/2) ÷ (4, 248.84/42) = 4.1768. The critical F value fromSAS is F (0.975,df =(2, 42)) = 4.0327. Since F∗> 4. 0327, conclude Ha: not both β2and β3= 0.We can obtain the p-value using SAS, wi t h the command: “pf=1-probf(4.1768,2,42)” . The p-value i s= 0.0221. Since the p-value i s less than 0.025, we conclude the alternative. In other words, at leastone of β2and β3are not equal to zero.Analysis of VarianceSum of MeanSource DF Squares Square F Value Pr > FModel 3 9120.46367 3040.15456 30.05 <.0001Error 42 4248.84068 101.16287Corrected Total 45 13369Root MSE 10.05798 R-Square 0.6822Dependent Mean 61.56522 Adj R-Sq 0.6595Coeff Var 16.33711Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t| Type I SSIntercept 1 158.49125 18.12589 8.74 <.0001 174353X1 1 -1.14161 0.21480 -5.31 <.0001 8275.38885X2 1 -0.44200 0.49197 -0.90 0.3741 480.91529X3 1 -13.47016 7.09966 -1.90 0.0647 364.15952Obs fcri1 fprob11 4.03271 0.0221614) 7.26a) The estimated regression function isˆY = 156.67 − 1.268X1− 0.921X2.Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t| Type I SSIntercept 1 156.67186 18.63964 8.41 <.0001 174353X1 1 -1.26765 0.21035 -6.03 <.0001 8275.38885X2 1 -0.92079 0.43489 -2.12 0.0401 480.91529b) For the mode l with all three variables X1,X2,X3, the estimated regression function isˆY = 158.49 −1.142X1− 0.442X2− 13.470X3. We see that the regression coefficients b0,b1,b2all change when addingthe third vari abl e X3.c) SSR(X1) = 8275.4, SSR(X1| X3) = 3483.9, so the y are not equal. SSR(X2) = 4860.3, SSR(X2|X3) = 708.0, so they are n ot equal .22Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t| Type I SSIntercept 1 119.94317 7.08475 16.93 <.0001 174353X1 1 -1.52060 0.17985 -8.45 <.0001 8275.38885Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t| Type I SSIntercept 1 145.94123 11.52509 12.66 <.0001 174353X3 1 -16.74205 6.08083 -2.75 0.0086 5554.91315X1 1 -1.20047 0.20411 -5.88 <.0001 3483.89147Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t| Type I SSIntercept 1 183.07696 24.32489 7.53 <.0001 174353X2 1 -2.40928 0.48059 -5.01 <.0001 4860.26000Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t| Type I SSIntercept 1 181.57256 22.49271 8.07 <.0001 174353X3 1 -25.14023 8.62949 -2.91 0.0057 5554.91315X2 1 -1.23948 0.59884 -2.07 0.0445 707.99714d) T he correlation matrix from SAS i s below. We see that the correlation of X1and X3is 0. 5697,and correlation of X2and X3is 0.6705. Sin ce these variables are correlated, the estim at ed regressioncoefficients and SSR values would change when adding X3to the model.Pearson Correlation Coefficients, N = 46Prob > |r| under H0: Rho=0X1 X2 X3X1 1.00000 0.56795 0.56968<.0001 <.0001X2 0.56795 1.00000 0.67053<.0001 <.0001X3 0.56968 0.67053 1.00000<.0001